[leetcode] 921. Minimum Add to Make Parentheses Valid

[leetcode] 921. Minimum Add to Make Parentheses Valid

Description

Given a string S of ‘(’ and ‘)’ parentheses, we add the minimum number of parentheses ( ‘(’ or ‘)’, and in any positions ) so that the resulting parentheses string is valid.

Formally, a parentheses string is valid if and only if:

It is the empty string, orIt can be written as AB (A concatenated with B), where A and B are valid strings, orIt can be written as (A), where A is a valid string.

Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.

Example 1:

Input: "())"Output: 1

Example 2:

Input: "((("Output: 3

Example 3:

Input: "()"Output: 0

Example 4:

Input: "()))(("Output: 4

Note:

S.length <= 1000S only consists of ‘(’ and ‘)’ characters.

分析

题目的意思是：给定括号字符串，求最小添加的括号数使得字符串合法。 我的思路很直接，一般括号是用栈来解决问题，所以我用栈stack来匹配合法的括号，res记录不合法的’)’，不合法的’('被栈记录上了，所以最后得到的结果为：res+len(stack)

代码

class Solution: def minAddToMakeValid(self, S: str) -> int: stack=[] res=0 for ch in S: if(ch=='('): stack.append(ch) elif(ch==')'): if(stack and stack[-1]=='('): stack.pop() continue else: res+=1 return res+len(stack)

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